Find the sum of the geometric series $1 -\dfrac45 + \left(\dfrac45\right)^2 -\left(\dfrac45\right)^3 +... -\left(\dfrac45\right)^{29}$ Choose 1 answer: Choose 1 answer: (Choice A) A $-0.03$ (Choice B) B $ 0.55 $ (Choice C) C $1.80$ (Choice D) D $5.01$
Solution: Getting started We're dealing with a geometric series because each term is multiplied by $-\dfrac45$ to get the next term. We need a formula to compute the sum of the terms. Formula for geometric series The sum $S_n$ of a finite geometric series is $S_n = \dfrac{a_1(1-r^n)}{1-r}$ where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. What do we need to use the formula? The first term $(a_1 = {1})$ is given in the question. The number of terms $n$ is ${30}$ because there are ${30}$ numbers from $0$ to $29$. [Where do the 0 and 29 come from?] The common ratio $r$ is ${-\dfrac45}$ because each term is multiplied by ${-\dfrac45}$ to get the next term. [How did we find the common ratio r?] Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac{a_1(1-r^n)}{1-r} \\\\ S_{{30}}&=\dfrac{{1}\left(1-\left({-\dfrac45}\right)^{{30}}\right)}{1-\left({-\dfrac45}\right)} \\\\ S_{{30}}&=\dfrac59\left(1-\left({-\dfrac45}\right)^{{30}}\right) \\\\ S_{{{30}}} &\approx 0.55 \end{aligned}$ The answer $ 0.55 $